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Rom4ik [11]
3 years ago
7

The nearest star to Earth is Proxima Centauri, 4.3 light-years away.

Physics
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

a) v = 2,152 10⁸ m / s   b)   t = 2.71 10⁸ s  or t =  85.93 year

Explanation:

a) In this special relativity exercise we have that time is measured in the same ship, so it is the proper time,

    v = d / t

Let's reduce the distance to the SI system

    d = 4.3 l and (9.46 1015 m / 1ly) = 40.678 10¹⁵ m

    t = 5.0 y (365 day / 1 y) (24 h / 1 day) (3600s / 1h) = 1.89 10⁸ s

Let's calculate

    v = 40.678 10¹⁵ / 1.89 10⁸

    v = 2,152 10⁸ m / s

b) The time seen from the ground for which the ship moves is given by

     t = t₀ / √ (1- (v/c)²)

Let's calculate

     t = 1.89 10⁸ / √ (1 - (2.152 / 2.998)²)

     t = 1.89 10⁸ / 0.6962

     t = 2.71 10⁸ s

Let's reduce this time to years

    t = 2.71 10⁸ s (1h / 3600s) (1day / 24h) (1 and / 365 d)

    t =  85.93 year

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PlzHELP
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 Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement<span> is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

</span>To calculate displacement<span>, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your </span>displacement<span> is 0. In physics, </span>displacement<span> is represented by Δs.
 
 For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me. 
 
https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php

Hope I helped. 

</span> 

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